Seg AM is a median of ∆ABC. If AB = 22, AC = 34, BC = 24, find AM
Chapter 2 – Pythagoras Theorem- Text Book Solution
Problem Set 2 | Q 18 | Page 46
Seg AM is a median of ∆ABC. If AB = 22, AC = 34, BC = 24, find AM
\[{AB}^2 + {AC}^2 = 2 {AM}^2 + 2 {BM}^2 \left( \text{by Apollonius theorem} \right)\]
\[ \Rightarrow \left( 22 \right)^2 + \left( 34 \right)^2 = 2 {AM}^2 + 2 \left( 12 \right)^2 \]
\[ \Rightarrow 484 + 1156 = 2 {AM}^2 + 288\]
\[ \Rightarrow 1640 - 288 = 2 {AM}^2 \]
\[ \Rightarrow 1352 = 2 {AM}^2 \]
\[ \Rightarrow {AM}^2 = 676\]
\[ \Rightarrow AM = 26\]
Hence, AM = 26.
Explanation:-
Given: $BM = MC = \frac{1}{2}BC = 12$
Using the Apollonius theorem in triangle $ABC$ with respect to median $AM$,
\begin{align*} {AB}^2 + {AC}^2 &= 2 {AM}^2 + 2 {BM}^2 \ \Rightarrow (22)^2 + (34)^2 &= 2 {AM}^2 + 2 (12)^2 \ \Rightarrow 484 + 1156 &= 2 {AM}^2 + 288 \ \Rightarrow 1640 &= 2 {AM}^2 \ \Rightarrow {AM}^2 &= 820 \ \Rightarrow AM &= \sqrt{820} = 2\sqrt{205} \end{align*}
Hence, $AM = 2\sqrt{205}$.
Chapter 2 – Pythagoras Theorem- Text Book Solution
Problem Set 2 | Q 18 | Page 46